Question

What are all the possible rational zeros for `f(x)=3x^3-8x^2-2x-3` and how do you find all zeros?

Answer 1

`f(x)` has zeros `3` and `-1/6+-sqrt(11)/6i`

Explanation:

`f(x) = 3x^3-8x^2-2x-3`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-3` and `q` a divisor of the coefficient `3` of the leading term.

That means that the only possible rational zeros are:

`+-1/3, +-1, +-3`

In addition note that the signs of the coefficients of `f(x)` are in the pattern `+ - - -`. By Descartes' Rule of signs, since there is one change of sign, `f(x)` has exactly one positive Real zero.

So check the possible positive rational zeros first:

`f(1/3) = 3(1/27)-8(1/9)-2(1/3)-3 = (1-8-6-27)/9= -40/9`

`f(1) = 3-8-2-3 = -10`

`f(3) = 3(27)-8(9)-2(3)-3 = 81-72-6-3 = 0`

So `x=3` is a zero and `(x-3)` a factor:

`3x^3-8x^2-2x-3 = (x-3)(3x^2+x+1)`

Use the quadratic formula to solve `3x^2+x+1=0` to find the two remaining zeros:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-1+-sqrt(1^2-4(3)(1)))/(2*3)`

`color(white)(x) = (-1+-sqrt(1-12))/6`

`color(white)(x) = (-1+-sqrt(-11))/6`

`color(white)(x) = -1/6+-sqrt(11)/6i`